Polyphase Networks - Massachusetts Institute of Technology

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Massachusetts Institute of Technology 
Department of Electrical Engineering and Computer Science 
6.061 Introduction to Power Systems 
Class Notes Chapter 3 
Polyphase Networks ∗ 
J.L. Kirtley Jr. 
1 Introduction 
Most electric power applications employ three phases. That is, three separate power carrying 
circuite, with voltages and currents staggered symmetrically in time are used. Two major reasons 
for the use of three phase power are economical use of conductors and nearly constant power flow. 
Systems with more than one phase are generally termed polyphase. Three phase systems are 
the most common, but there are situations in which a different number of phases may be used. 
Two phase systems have a simplicity that makes them useful for teaching vehicles and for certain 
servomechanisms. This is why two phase machines show up in laboratories and textbooks. Sys­
tems with a relatively large number of phases are used for certain specialized applications such as 
controlled rectifiers for aluminum smelters. Six phase systems have been proposed for very high 
power transmission applications. 
Polyphase systems are qualitatively different from single phase systems. In some sense, polyphase 
systems are more complex, but often much easier to analyze. This little paradox will become ob­
vious during the discussion of electric machines. It is interesting to note that physical conversion 
between polyphase systems of different phase number is always possible. 
This chapter starts with an elementary discussion of polyphase networks and demonstrates 
some of their basic features. It ends with a short discussion of per-unit systems and power system 
representation. 
2 Two Phases 
The two-phase system is the simplest of all polyphase systems to describe. Consider a pair of 
voltage sources sitting side by side with: 
v1 = V cosωt (1) 
v2 = V sinωt (2) 
∗ c�2003 James L. Kirtley Jr. 
1
�� 
�� 
�� 
+ �� 
� � 
i1 i2 
+ 
+ + v2v1 v2 
− 
v1 Z − Z 
− − 
jv1 = Re 
� 
ωt V e
� 
(3) � 
jωtv2 = Re − jV e
� 
(4) 
= 
� 
)Re j(ωt
π 
V e − 2 
� 
(5) 
� V1 
� 
Suppose this system of sources is connected to al “balanced load”, as shown in Figure 1. To 
compute the power flows in the system, it is convenient to re-write the voltages in complex form: 
Figure 1: Two-Phase System 
V2 
Figure 2: Phasor Diagram for Two-Phase Source
If each source is connected to a load with impedance:
Z = Z ejψ
| |
then the complex amplitudes of currents are: 
V 
−jψ I1 = e |Z|
V 
−jψ −j π I2 = e e 2 |Z| 
Each of the two phase networks has the same value for real and reactive power: 
P + jQ = 
|V |2 
ejψ (6) 
2 Z| | 
2 
� � 
� � 
3 
or: 
P = 
|V |2 
cosψ (7) 
2 Z| | 
Q = 
|V |2 
sinψ (8) 
2 Z| | 
The relationship between “complex power” and instantaneous power flow was worked out in 
Chapter 2 of these notes. For a system with voltage of the form: 
v = Re V ejφejωt 
instantaneous power is given by: 
p = P [1 + cos 2(ωt + φ)] +Q sin 2(ωt + φ) 
For the case under consideration here, φ = 0 for phase 1 and φ = π 2 for phase 2. Thus: −
p1 = 
|V |2 
cosψ [1 + cos 2ωt] +
|V |2 
sinψ sin 2ωt 
2 Z 2 Z| | | | 
p2 = 
|V |2 
cosψ [1 + cos(2ωt − π)] + |V |
2 
sinψ sin(2ωt − π)
2 Z 2 Z| | | | 
Note that the time-varying parts of these two expressions have opposite signs. Added together, 
they give instantaneous power: 
p = p1 + p2 = 
|V |2 
cosψ |Z| 
At least one of the advantages of polyphase power networks is now apparent. The use of a 
balanced polyphase system avoids the power flow pulsations due to ac voltage and current, and 
even the pulsations due to reactive energy flow. This has obvious benefits when dealing with 
motors and generators or, in fact, any type of source or load which would like to see constant 
power. 
Three Phase Systems 
Now consider the arrangement of three voltage sources illustrated in Figure 3. 
The three phase voltages are: 
va = V cosωt = Re V e
jωt (9) 
)vb = V cos(ωt − 23 π ) = Re 
� 
V ej(ωt−
2
3 
π 
� 
(10) 
vc = V cos(ωt + 
2
3 
π 
� 
2π 
� 
) = Re V ej(ωt+ 3 ) (11) 
These three phase voltages are illustrated in the time domain in Figure 4 and as complex 
phasors in Figure 5. Note the symmetrical spacing in time of the voltages. As in earlier examples, 
the instantaneous voltages may be visualized by imagining Figure 5 spinning counterclockwise with 
3
� � 
� � 
� � 
+ 
+ 
�� �� ��+ vc vb+ + + 
�� V ej 
2
3 �� V e−j 
2
3�� V va − − − 
π π 
− − − 
Figure 3: Three- Phase Voltage Source
−1 
−0.8 
−0.6 
−0.4 
−0.2 
0 
0.2 
0.4 
0.6 
0.8 
1 
Va
, V
b,
 V
c 
0 5 10 15 
t 
Figure 4: Three Phase Voltages
angular velocity ω. The instantaneous voltages are just projections of the vectors of this “pinwheel” 
onto the horizontal axis. 
Consider connecting these three voltage sources to three identical loads, each with complex 
impedance Z, as shown in Figure 6. 
If voltages are as given by (9 - 11), then currents in the three phases are: 
V 
ia = Re e
jωt (12) 
Z 
V j(ωt− 2e 3 
π 
ib = Re 
) (13) 
Z
V j(ωt+ 2e 3 
π 
ic = Re 
) (14) 
Z 
4 
�� �� �� 
� � � 
ic ib ia 
�� 
v +a
�� 
+ +
 
vb
�� + + + 
vc
− − − 
 
vc vb vZ
 Z
 Z
 a
�in 
− − − 
� 
� 
� 
�
� �
� 
� 
� 
� 
�
Vc 
� �Va 
Vb ��
Figure 5: Phasor Diagram: Three Phase Voltages 
Figure 6: Three- Phase Source Connected To Balanced Load 
Complex power in each of the three phases is: 
P + jQ = 
|V |2 
(cosψ + j sinψ) (15)
2 Z| | 
Then, remembering the time phase of the three sources, it is possible to write the values of instan­
taneous power in the three phases: 
pa = 
|V |2 {cosψ [1 + cos 2ωt] + sinψ sin 2ωt} (16)
2 Z|
2
| � � � � 
pb = 
|V |
cosψ 1 + cos(2ωt − 2π 2π |
2
| � � 
) 
� 
+ sinψ sin(2ωt − 
� 
) (17)
2 Z 3 3 
pc = 
|V |
cosψ 1 + cos(2ωt +
2π 2π 
) + sinψ sin(2ωt + ) (18)
2 Z 3 3| | 
The sum of these three expressions is total instantaneous power, which is constant: 
3 V 2 
p = pa + pb + pc = 
| |
cosψ (19)
2 Z| | 
5 
 
�� 
�
�� 
�� 
�
 
Z 
vb ib + vb 
− + 
−
Z
vc ic + vc 
�− + 
−
� Z 
� � � � 
��
�� + �
� � � � 
� � � � 
� � � � 
It is useful, in dealing with three phase systems, to remember that 
2π 2π 
cosx + cos(x − ) + cos(x + ) = 0 
3 3 
regardless of the value of x. 
Now consider the current in the neutral wire, in in Figure 6. This current is given by: 
V 
in = ia + ib + ic = Re e
jωt +
 e
j(ωt−
2π 2π) j(ωt++ ) =
 0 (20)
3 3e
Z
This shows the most important advantage of three-phase systems over two-phase systems: a 
wire with no current in it does not have to be very large. In fact, the neutral connection may 
be eliminated completely in many cases. The network shown in Figure 7 will work as well as the 
network in Figure 6 in most cases in which the voltages and load impedances are balanced. 
va ia 
− +
va −
Figure 7: Ungrounded Three-Phase Source and Load 
There is a fundamental difference between grounded and undgrounded systems if perfectly 
balanced conditions are not maintained. In effect, the ground wire provides isolation between the 
phases by fixing the neutral voltage a the star point to be zero. If the load impedances are not 
equal the load is said to be unbalanced. If the system is grounded there will be current in the 
neutral. If an unbalanced load is not grounded, the star point voltage will not be zero, and the 
voltages will be different in the three phases at the load, even if the voltage sources all have the 
same magnitude. 
Line-Line VoltagesA balanced three-phase set of voltages has a well defined set of line-line voltages. If the line-to­
neutral voltages are given by (9 - 11), then line-line voltages are: 
−j 2
3 
π jωt vab = va − vb = Re V
 1−
 e
 (21)
e
−j 2π j 2π jωt (22) Re V
 3 3 
− 1 ejωt (23) 
vbc = vb − vc =
 −
e
 e
 e
j 2
3 
π 
Re V
vca = vc − va = e
6
4 
 
and these reduce to:
j π jωt vab = Re
√�
3V e 6 e
� 
(24)
= 
√�
3 −j
π 
2 
jωt	vbc Re V e e
� 
(25) 
= 
√�
3 j 
5π
6 
jωt	vca Re V e e
� 
(26) 
The phasor relationship of line-to-neutral and line-to-line voltages is shown in Figure 8. Two things 
should be noted about this relationship: 
•	 The line-t√o-line voltage set has a magnitude that is larger than the line-ground voltage by a 
factor of 3. 
•	 Line-to-line voltages are phase shifted by 30◦ ahead of line-to-neutral voltages. 
Clearly, line-to-line voltages themselves form a three-phase set just as do line-to-neutral voltages. 
Power system components (sources, transformer windings, loads, etc.) may be connected either 
between lines and neutral or between lines. The former connection of often called wye, the latter 
is called delta, for obvious reasons.
Vca 
Vc 
Va 
Vab 
Vb 
Vbc 
Vbc 
VabVca 
Va 
Vb 
Vc 
Figure 8: Line-Neutral and Line-Line Voltages 
It should be noted that the wye connection is at least potentially a four-terminal connection, 
while the delta connection is inherently three-terminal. The difference is the availability of a neutral 
point. Under balanced operating conditions this is unimportant, but the difference is apparent and 
important under unbalanced conditions. 
4.1 Example: Wye and Delta Connected Loads 
Loads may be connected in either line-to-neutral or line-to-line configuration. An example of the 
use of this flexibility is in a fairly commonly used distribution system with a line-to-neutral voltage 
of 120 V, RMS. In this system the line-to-line voltage is 208 V, RMS. Single phase loads may be 
connected either line-to-line or line-to-neutral. 
7 
- -
+ 
Va - + 
- Vca Vab 
Vb + 
- + 
V+ c - V + bc 
Figure 9: Wye And Delta Connected Voltage Sources 
a a 
Z
a 
ca 
Z Z
ab 
Z b 
Z c b c Z bc bc 
Figure 10: Wye And Delta Connected Impedances 
Suppose it is necessary to build a resistive heater to deliver 6 kW, to be made of three elements 
which may be connected in either wye or delta. Each of the three elements must dissipate 2000 W. 
Thus, since P = V
R 
2 
, the wye connected resistors would be: 
1202 
Ry = = 7.2Ω 
2000 
while the delta connected resistors would be: 
2082 
RΔ = = 21.6Ω 
2000 
As is suggested by this example, wye and delta connected impedances are often directly equiv­
alent. In fact, ungrounded connections are three-terminal networks which may be represented in 
two ways. The two networks shown in Figure 10, combinations of three passive impedances, are 
directly equivalent and identical in their terminal behavior if the relationships between elements 
are as given in (27 - 32). 
Zab 
ZaZb + ZbZ + Z Zc c a = (27) 
Zc 
8 
 
ZaZb + ZbZc + Z Za Zbc = 
c (28) 
Za 
ZaZb + ZbZ
c = 
c + ZcZaZ a (29) 
Zb 
ZabZ a Za = 
c (30) 
Zab + Zbc + Zca 
ZabZbc Zb = (31) 
Zab + Zbc + Zca 
Zb Zca Z = cc (32) 
� � 
� � 
� � 
Zab + Zbc + Zca 
A special case of the wye-delta equivalence is that of balanced loads, in which: 
Za = Zb = Zc = Zy 
and 
Zab = Zbc = Zca = ZΔ 
in which case: 
ZΔ = 3Zy 
4.2 Example: Use of Wye-Delta for Unbalanced Loads 
The unbalanced load shown in Figure 11 is connected to a balanced voltage source. The problem 
is to determine the line currents. Note that his load is ungrounded (if it were grounded, this would 
be a trivial problem). The voltages are given by: 
va = V cosωt 
2π 
vb = V cos(ωt − )
3 
2π 
vc = V cos(ωt + )
3 
To solve this problem, convert both the source and load to delta equivalent connections, as 
shown in Figure 12. The values of the three resistors are: 
2 + 4 + 2 
rab = rca = = 4 
2 
2 + 4 + 2 
rbc = = 8 
1
The complex amplitudes of the equivalent voltage sources are:
V 
√
3e
−j 
2
3 
j π 
6 
π 
V ab = V a − V b = V 1− e
 =
V 
√
3e
π 
2
−j 2
3 
j 2
3 
π π
−jV bc = V b − V c = V −
 e
 =
e
V 
√
3e
j 
5
6 
πj 2
3 
π − 1
V ca = V c − V a = V =
e
9
 
Ia 
+ 
Va R a= 1 
-
- -Vc Vb R b = 2 R c = 2 
+ + I 
I 
b 
 
 
 
c 
Figure 11: Unbalanced Load 
Ib 
Ia 
Ic 
R = 4
ab
R = 4 
ca 
R = 8bcVbc 
Vca V
ab 
- + 
-
+ 
-
+ 
Figure 12: Delta Equivalent
10
� � 
� � � � 
� � � � 
� � 
� 
��� 
�
�
�
� 
��� 
5 
Currents in each of the equivalent resistors are: 
V ab V bc V caI1 = I2 = I3 = 
rcarab rbc 
The line curents are then just the difference between current in the legs of the delta: 
π 5π 3
je√3V je6 6Ia = I1 − I3 V
−
=
 =
4 4 4 
j π 
6e
√
3V 
−j π 
8 
2 −
 3 1
eIb = I2 − I1 
8
+ j V 
4
4 
= −=
5π π 3 1je√3V −j6 2eIc = I3 − I2 =
 −
8 
− j 
4 
V
4 −
=
 8 
These are shown in Figure 13. 
Im( )·
1V4
33V 
�
� V8 4−
Re( )·
1V4−
Figure 13: Line Currents 
Transformers 
Transformers are essential parts of most power systems. Their role is to convert electrical energy 
at one voltage to some other voltage. We will deal with transformers as electromagnetic elements 
later on in this subject, but for now it will be sufficient to use a simplified model for the transformer 
which we will call the ideal transformer. This is a two-port circuit element, shown in Figure 14. 
i1 i2� N1 : N2 �
+ + 
v1 v2 
− ⊃
⊃⊃
⊃
⊂⊂
⊂⊂
− 
Figure 14: Ideal Transformer
The ideal transformer as a network element constrains its terminal variables in the following 
way: 
11 
I�1 N1 : N2 
I�2 
+ 
V 1 
− ⊃
⊃⊃
⊃
⊂⊂
⊂⊂ Z 
v1 v2 
= (33)
N1 N2 
N1i1 = −N2i2 (34) 
As it turns out, this is not a terribly bad model for the behavior of a real transformer under 
most circumstances. Of course, we will be interested in fine points of transformer behavior and 
behavior under pathological operating conditions, and so will eventually want a better model. For 
now, it is sufficient to note just a few things about how the transformer works. 
1. In normal operation, we select a transformer turns ratio N1 so that the desired voltages
N2 
appear at the proper terminals. For example, to convert 13.8 kV distribution voltage to the 
120/240 volt level suitable for residential or commercial single phase service, we would use a 
transformer with turns ratio of 13800 = 57.5. To split the low voltage in half, a center tap on240 
the low voltage winding would be used. 
2. The transformer, at least in its ideal form, does not consume, produce nor store energy. Note 
that, according to (33) and (34), the sum of power flows into a transformer is identically zero: 
p1 + p2 = v1i1 + v2i2 = 0 (35) 
3. The transformer also tends to transform impedances. To show how this is, look at Figure 15. 
Here, some impedance is connected to one side of an ideal transformer. See that it is possible 
to find an equivalent impedance viewed from the other side of the transformer. 
Figure 15: Impedance Transformation 
Noting that 
N1
I2 = −
N2 
I1 
and that 
V 2 = −ZI2 
Then the ratio between input voltage and current is: 
V 1 = 
N1 
N2 
V 2 = 
� 
N1 
N2 
�2 
I1 (36) 
12 
� 
� 
� 
� � 
� 
� 
� 
⊃⊃
⊃⊃
⊂⊂
⊂⊂
�
� 
⊃⊃
⊃⊃
⊂⊂
⊂⊂
�
� 
⊃⊃
⊃⊃
⊂⊂
⊂⊂
6 Three-Phase Transformers 
A three-phase transformer is simply three single phase transformers.The complication in these 
things is that there are a number of ways of winding them, and a number of ways of interconnecting 
them. We will have more to say about windings later. For now, consider interconnections. On 
either “side” of a transformer connection (i.e. the high voltage and low voltage sides), it is possible 
to connect transformer windings either line to neutral (wye), or line to line (delta). Thus we may 
speak of transformer connections being wye-wye, delta-delta, wye-delta, or delta-wye. 
Ignoring certain complications that we will have more to say about shortly, connection of trans­
formers in either wye-wye or delta-delta is reasonably easy to understand. Each of the line-to-neutral 
(in the case of wye-wye), or line-to-line (in the case of delta-delta) voltages is transformed by one 
of the three transformers. On the other hand, the interconnections of a wye-delta or delta-wye 
transformer are a little more complex. Figure 16 shows a delta-wye connection, in what might be 
called “wiring diagram” form. A more schematic (and more common) form of the same picture is 
shown in Figure 17. In that picture, winding elements that appear parallel are wound on the same 
core segment, and so constitute a single phase transformer. 
Xc Hc Xb Hb Xa Ha 
Figure 16: Delta-Wye Transformer Connection 
Now: assume that NΔ and NY are numbers of turns. If the individual transformers are consid­
ered to be ideal, the following voltage and current constraints exist: 
NY 
vaY = 
NΔ 
(vaΔ − vbΔ) (37) 
vbY = 
NY 
NΔ 
(vbΔ − vcΔ) (38) 
vcY = 
NY 
NΔ 
(vcΔ − vaΔ) (39) 
iaΔ = 
NY 
NΔ 
(iaY − icY ) (40) 
ibΔ = 
NY 
NΔ 
(ibY − iaY ) (41) 
icΔ = 
NY 
NΔ 
(icY − ibY ) (42) 
13 
� � 
� � 
� � 
� � 
� � 
� � 
AYA 
N 
NY 
C N 
NY 
N 
NY 
B CY BY 
Figure 17: Schematic of Delta-Wye Transformer Connection 
where each of the voltages are line-neutral and the currents are in the lines at the transformer 
terminals. 
Now, consider what happens if a Δ − Y transformer is connected to a balanced three- phase 
voltage source, so that: 
vaΔ = Re V e
jωt 
)vbΔ = Re V e
j(ωt− 2
3 
π 
)vcΔ = Re V e
j(ωt+ 2
3 
π 
Then, complex amplitudes on the wye side are: 
= 1−	
NΔ 
V e 6V aY 
NY V e −j 
2
3 
π 
= 
√
3 
NY j π 
NΔ 
= NY V e 3 e 3 = 
√
3 
NY −j π V e 2V bY NΔ 
−j 2π − j 2π 
NΔ 
NY j 5π = NY V ej 
2
3 
π − 1 = 
√
3 
NΔ 
V e 6V cY NΔ 
Two observations should be made here: 
•	 The ratio of voltages (that is, the ratio of either line-line or line-neutral) is different from the 
turns ratio by a factor of 
√
3. 
•	 All wye side voltages are shifted in phase by 30◦ with respect to the delta side voltages. 
6.1 Example 
Suppose we have the following problem to solve: 
14 
A balanced three- phase wye-connected resistor is connected to the Δ side of a Y −Δ 
transformer with a nominal voltage ratio of 
vΔ 
= N 
vY 
What is the impedance looking into the wye side of the transformer, assuming drive 
with a balanced source? 
The situation is shown in Figure 18. 
NY 
NY 
NY 
N 
N 
N R R 
R 
Figure 18: Example
It is important to remember the relationship between the voltage ratio and the turns ratio,
which is: 
vΔ 
vY 
= N = 
NΔ √
3NY 
so that: 
Nδ N 
NY 
= √
3 
Next, the Y −Δ equivalent transform for the load makes the picture look like figure 19 
In this situation, each transformer secondary winding is conected directly across one of the three 
resistors. Currents in the resistors are given by: 
vabΔ
i1 = 
3R 
vbcΔ
i2 = 
3R 
vcaΔ
i3 = 
3R 
Line currents are:
iaΔ = 
ibΔ = 
icΔ = 
i1 − i3 = vabΔ−vcaΔ 3R 
i2 − i1 = vbcΔ−vabΔ 3R 
i3 − i2 = vcaΔ−vbcΔ 3R 
= i1Δ − i3Δ 
= i2Δ − i1Δ 
= i3Δ − i2Δ 
15 
NΔ NΔ 
vabΔ = vaY iaY = i1Δ 
NY NY 
NΔ NΔ
vbcΔ = vbY ibY = i2Δ 
NY NY 
NΔ NΔ 
vcaΔ = vcY icY = i3Δ 
NY NY 
� �2NΔ 1 
iaY = vaY 
NY 3R 
16
N 
N 
N 
N Y 
N Y 
NY 
+ 
− 
+ 
−
 
+ 
−
vbY 
i
aY 
i
cY 
+ 
ca 
vab 
i
v
aY 
vcY 
ibY 
+ 
−
 
+ 
−
−
 
v
vbc 
i a 
ib 
i c 
ii 2 
3 
3R 
3R 
3R 
1 
Figure 19: Equivalent Situation 
Solving for currents in the legs of the transformer Δ, subtract, for example, the second expression 
from the first: 
2i1Δ − i2Δ − i3Δ = 2vabΔ − vbcΔ − vcaΔ 
3R 
Now, taking advantage of the fact that the system is balanced: 
i1Δ + i2Δ + i3Δ = 0 
vabΔ + vbcΔ + vcaΔ = 0 
to find: 
vabΔ
i1Δ = 
3R 
vbcΔ
i2Δ = 
3R 
vcaΔ
i3Δ = 
3R 
Finally, the ideal transformer relations give: 
so that:
1 
= 
�
Δ 
�2N
ibY vbY 
NY 3R � �2NΔ 1 
icY = vcY 
NY 3R 
The apparent resistance (that is, apparent were it to be connected in wye) at the wye terminals 
of the transformer is: � �2NY
Req = 3R 
NΔ 
Expressed in terms of voltage ratio, this is: 
� �2 N
= 
�
v
Δ 
�2
Y
Req = 3R √ R
3 v
It is important to note that this solution took the long way around. Taken consistently (uni­
formly on a line-neutral or uniformly on a line-line basis), impedances transform across transformers 
by the square of the voltage ratio, no matter what connection is used. 
7 Polyphase Lines and Single-Phase Equivalents 
By now, one might suspect that a balanced polyphase system may be regarded simply as three 
single-phase systems, even though the three phases are physically interconnected. This feeling is 
reinforced by the equivalence between wye and delta connected sources and impedances. One more 
step is required to show that single phase equivalence is indeed useful, and this concerns situations 
in which the phases have mutual coupling. 
In speaking of lines, we mean such system elements as transmission or distribution lines: over­
head wires, cables or even in-plant buswork. Such elements have impedance, so that there is some 
voltage drop between the sending and receiving ends of the line. This impedance is more than just 
conductor resistance: the conductors have both self and mutual inductance, because currents in 
the conductors make magnetic flux which, in turn, is linked by all conductors of the line. 
A schematic view of a line is shown in Figure 20. Actually, only the inductance components of 
line impedance are shown, since they are the most interesting parts of line impedance. Working in 
complex amplitudes, it is possible to write the voltage drops for the three phases by: 
V a1 − V a2 = jωLIa + jωM (Ib + Ic) (43) 
V b1 − V b2 = jωLIb + jωM (Ia + Ic) (44) 
V c1 − V c2 = jωLIc + jωM (Ia + Ib) (45) 
If the currents form a balanced set: 
Ia + Ib + Ic = 0 (46) 
Then the voltage drops are simply: 
V a1 − V a2 = jω (L −M) Ia 
V b1 − V b2 = jω (L −M) Ib 
V c1 − V c2 = jω (L −M) Ic 
17 
� � 
�
�
L
M 
M M 
L 
L 
v 
v 
v 
a2 
b2 
c2c1
v
b1 
v 
a1v
i 
i 
a 
b 
i 
c 
Figure 20: Schematic Of A Balanced Three-Phase Line With Mutual Coupling 
- Vb 
+ 
Va 
-
+ 
V 
c 
-
+ 
N 
N 
N 
M 
M M 
L 
L 
L 
NY 
N Y 
NY 
R 
R 
R 
Voltage Source Primary Secondary Transmission Line Load 
Transformer 
Figure 21: Example 
In this case, an apparent inductance, suitable for the balanced case, has been defined: 
L1 = L −M (47) 
which describes the behavior of one phase in terms of its own current. It is most important to note 
that this inductance is a valid description of the line only if (46) holds, which it does, of course, in 
the balanced case. 
7.1 Example 
To show how the analytical techniques which come from the network simplification resulting from 
single phase equivalents and wye-delta transformations, considerthe following problem: 
A three-phase resistive load is connected to a balanced three-phase source through a 
transformer connected in delta-wye and a polyphase line, as shown in Figure 21. The 
problem is to calculate power dissipated in the load resistors. The three- phase voltage 
source has: 
jωt va = Re 
√
2VRMSe
j(ωt− 2π 
� 
)vb = Re 
√
2VRMSe 3 
j(ωt+ 2π 
� 
)vc = Re 
√
2VRMSe 3 
18 
This problem is worked by a succession of simple transformations. First, the delta connected 
resistive load is converted to its equivalent wye with RY = 
R 
3 . 
Next, since the problem is balanced, the self- and mutual inductances of the line are directly 
equivalent to self inductances in each phase of L1 = L −M . 
Now, the transformer secondary is facing an impedance in each phase of: 
ZY s = jωL1 + RY 
The delta-wye transformer has a voltage ratio of: 
vp NΔ 
= 
vs 
√
3NY 
so that, on the primary side of the transformer, the line and load impedance is: 
Zp = jωLeq + Req 
where the equivalent elements are: 
1 
� �2NΔ 
(L −M)Leq = 
3 NY 
1 
� 
NΔ 
�2 R 
Req = 
3 NY 3 
Magnitude of current flowing in each phase of the source is: 
√
2VRMS |I| = � 
)2 + R2(ωLeq eq 
Dissipation in one phase is: 
1 2P1 = 
2
|I| Req 
V 2 RMSReq = 
(ωLeq)
2 + Req 
2 
And, of course, total power dissipated is just three times the single phase dissipation. 
Introduction To Per-Unit Systems 
Strictly speaking, per-unit systems are nothing more than normalizations of voltage, current, 
impedance and power. These normalizations of system parameters because they provide sim­
plifications in many network calculations. As we will discover, while certain ordinary parameters 
have very wide ranges of value, the equivalent per-unit parameters fall in a much narrower range. 
This helps in understanding how certain types of system behave. 
19 
8 
�� 
�� 
� 
i1 
+ 
+ 
V v
− Z 1
− 
Figure 22: Example 
8.1 Normalization Of Voltage And Current 
The basis for the per-unit system of notation is the expression of voltage and current as fractions 
of base levels. Thus the first step in setting up a per-unit normalization is to pick base voltage and 
current. 
Consider the simple situation shown in Figure 22. For this network, the complex amplitudes 
of voltage and current are: 
V = IZ (48) 
We start by defining two base quantities, VB for voltage and IB for current. In many cases, these 
will be chosen to be nominal or rated values. For generating plants, for example, it is common to 
use the rated voltage and rated current of the generator as base quantities. In other situations, 
such as system stability studies, it is common to use a standard, system wide base system. 
The per-unit voltage and current are then simply: 
V 
v = (49) 
VB 
I 
i = (50) 
IB 
Applying (49) and (50) to (48), we find: 
v = iz (51) 
where the per-unit impedance is: 
IB 
z = Z (52) 
VB 
This leads to a definition for a base impedance for the system: 
VB
ZB = (53) 
IB 
Of course there is also a base power, which for a single phase system is: 
PB = VBIB (54) 
as long as VB and IB are expressed in RMS. It is interesting to note that, as long as normalization is 
carried out in a consistent way, there is no ambiguity in per-unit notation. That is, peak quantities 
normalized to peak base quantities will be the same, in per-unit, as RMS quantities normalized to 
RMS bases. This advantage is even more striking in polyphase systems, as we are about to see. 
20
8.2 Three Phase Systems 
When describing polyphase systems, we have the choice of using either line-line or line-neutral 
voltage and line current or current in delta equivalent loads. In order to keep straight analysis in 
ordinary variable, it is necessary to carry along information about which of these quantities is being 
used. There is no such problem with per-unit notation. 
We may use as base quantities either line to neutral voltage VBl−g or line to line voltage VBl−l. 
Taking the base current to be line current IBl, we may express base power as: 
PB = 3VBl−gIBl (55) 
Because line-line voltage is, under normal operation, 
√
3 times line-neutral voltage, an equivalent 
statement is: 
PB = 
√
3VBl−lIBl (56) 
If base impedance is expressed by line-neutral voltage and line current (This is the common 
convention, but is not required), 
VBl−g
ZB = (57)
IBl 
Then, base impedance is, written in terms of base power: 
V 2 V 2PB Bl−g Bl−lZB =
3I2 
= 3 
PB 
= 
PB 
(58) 
B 
Note that a single per-unit voltage applied equally well to line-line, line-neutral, peak and RMS 
quantities. For a given situation, each of these quantities will have a different ordinary value, but 
there is only one per-unit value. 
8.3 Networks With Transformers 
One of the most important advantages of the use of per-unit systems arises in the analysis of 
networks with transformers. Properly applied, a per-unit normalization will cause nearly all ideal 
transformers to dissapear from the per-unit network, thus greatly simplifying analysis. 
To show how this comes about, consider the ideal transformer as shown in Figure 23. The 
�I1 
1 : N 
�I2 
+ + 
V 1 V 2 
− ⊃
⊃⊃
⊃
⊂⊂
⊂⊂
− 
Figure 23: Ideal Transformer With Voltage And Current Conventions Noted 
ideal transformer imposes the constraints that: 
V 2 = NV 1 
1 
I2 = 
N 
I1 
21 
Normalized to base quantities on the two sides of the transformer, the per-unit voltage and 
current are: 
V 1 v1 = 
VB1 
I1i1 = 
IB1 
V 2 v2 = 
VB2 
I2i2 = 
IB2 
Now: note that if the base quantities are related to each other as if they had been processed by the 
transformer: 
VB2 =	 NVB1 (59) 
IB1 
(60) IB2 = 
N 
then v1 = v2 and i1 = i2, as if the ideal transformer were not there (that is, consisted of an ideal 
wire). 
Expressions (59) and (60) reflect a general rule in setting up per-unit normalizations for systems 
with transformers. Each segment of the system should have the same base power. Base voltages 
transform according to transformer voltage ratios. For three-phase systems, of course, the voltage 
ratios may differ from the physical turns ratios by a factor of 
√
3 if delta-wye or wye-delta connections 
are used. It is, however, the voltage ratio that must be used in setting base voltages. 
8.4 Transforming From One Base To Another 
Very often data such as transformer leakage inductance is given in per-unit terms, on some base 
(perhaps the units rating), while in order to do a system study it is necessary to express the same 
data in per-unit in some other base (perhaps a unified system base). It is always possible to do this 
by the two step process of converting the per-unit data to its ordinary form, then re-normalizing it 
in the new base. However, it is easier to just convert it to the new base in the following way. 
Note that impedance in Ohms (ordinary units) is given by: 
Z = z1ZB1 = z2ZB2	 (61) 
Here, of course, z1 and z2 are the same per-unit impedance expressed in different bases. This could 
be written as: 
V 2 V 2 B1 B2 z1 = z2	 (62) 
PB1 PB2 
This yields a convenient rule for converting from one base system to another:
PB1 
� 
VB2 
�2 
z1 = z2 (63) 
PB2 VB1 
22
�� 
�� 
� 
xg xT 1 xl rl xT 2 
∨ ∨ 
∩∩∩∩ ∩∩∩∩ ∩∩∩∩ ∧ ∧ ∧ ∩∩∩∩ � 
+ eg
− 
T 
Load 
11 
G L T2 
3-Phase 
Fault 
Figure 24: One-Line Diagram Of Faulted System 
8.5 Example: Fault Study 
To illustrate some of the concepts with which we have been dealing, we will do a short circuit analysis 
of a simple power system. This system is illustrated, in one-line diagram form, in Figure 24. 
A one-line diagram is a way of conveying a lot of information about a power system withoutbecoming cluttered with repetitive pieces of data. Drawing all three phases of a system would 
involve quite a lot of repetition that is not needed for most studies. Further, the three phases can 
be re-constructed from the one-line diagram if necessary. It is usual to use special symbols for 
different components of the network. For our network, we have the following pieces of data: 
Symbol Component Base P Base V Impedance 
(MVA) (kV) (per-unit) 
G1 Generator 200 13.8 j.18 
T1 Transformer 200 13.8/138 j.12 
L1 Trans. Line 100 138 .02 + j.05 
T2 Transformer 50 138/34.5 j.08 
A three-phase fault is assumed to occur on the 34.5 kV side of the transformer T2. This is 
a symmetrical situation, so that only one phase must be represented. The per-unit impedance 
diagram is shown in Figure 25. It is necessary to proceed now to determine the value of the 
components in this circuit. 
Figure 25: Impedance Diagram For Fault Example 
First, it is necessary to establish a uniform base an per-unit value for each of the system 
components. Somewhat arbitrarily, we choose as the base segment the transmission line. Thus all 
of the parameters must be put into a base power of 100 MVA and voltage bases of 138 kV on the 
23
• At the generator |If | = 11, 595A 
• On the transmission line |If | = 1159A 
• At the fault |If | = 4633A 
line, 13.8 kV at the generator, and 34.5 kV at the fault. Using (62): 
100 xg =	 200 × .18 = .09per-unit 
100 xT 1 =	 200 × .12 = .06per-unit 
100 xT 2 = 50 × .08 = .16per-unit 
rl = = .02per-unit 
xl = = .05per-unit 
Total impedance is: 
z = j (xg + xT 1 + xl + xT 2) + rl 
= j.36 + .02per-unit 
|z| = .361per-unit 
Now, if eg is equal to one per-unit (generator internal voltage equal to base voltage), then the 
per-unit current is: 
1 |i| = 
.361 
= .277per-unit 
This may be translated back into ordinary units by getting base current levels. These are: 
•	 On the base at the generator:
100MVA
IB = √
3× 13.8kV = 4.18kA 
On the line base: •	
100MVA 
IB = √
3× 138kV = 418A
On the base at the fault:
•	
100MVA 
IB = √
3× 34.5kV = 1.67kA 
Then the actual fault currents are: 
24
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